heureka

2/4-1/4 (simplyfi)

\(\begin{array}{lcl} \frac24-\frac14 \\ &=& 2\cdot \frac14-\frac14 \\ &=& \frac14+\frac14-\frac14 \\ &=& \frac14 \\ \end{array}\)

-8 1/2 - (-3 1/4)

\(\begin{array}{lcr} -8 \frac12 - (-3 \frac14 )\\ &=& -8 \frac12 + 3 \frac14 \\ &=& -8 -\frac12 +3 + \frac14 \\ &=& -8 +3 - \frac12 + \frac14 \\ &=& -5 - \frac12 + \frac14 \\ &=& -5 - \frac24 + \frac14 \\ &=& -5 - \frac14 \\ &=& -5 \frac14 \\ -8 \frac12 - (-3 \frac14 )&=& -5 \frac14 \\ \end{array}\)

x + y + z = 11

0.70x + 0.55y + 0.50z = 6.70

x = 3z

\(\begin{array}{lrcrcrcr} (1) & x &+& y &+& z &=& 11 \\ (2) & 0.70x &+& 0.55y &+& 0.50z &=& 6.70 \\ \\ \hline \\ x= 3z: \\ \\ (1) & 3z &+& y &+& z &=& 11 \\ (2) & 0.70\cdot (3z) &+& 0.55y &+& 0.50z &=& 6.70 \\ \\ \hline \\ (1) & && y &+& 4z &=& 11 \\ (2) & 2.1\cdot z &+& 0.55y &+& 0.50z &=& 6.70 \\ \\ \hline \\ (1) & && y &+& 4z &=& 11 \\ (2) & && 0.55y &+& 2.60z &=& 6.70 \\ \\ \hline \\ (1) & && y && &=& 11 - 4z\\ (2) & && 0.55y &+& 2.60z &=& 6.70 \\ & && 0.55(11 - 4z) &+& 2.60z &=& 6.70 \\ & && 0.55\cdot 11 - 0.55\cdot 4z &+& 2.60z &=& 6.70 \\ & && 6.05 - 2.20z &+& 2.60z &=& 6.70 \\ & && 6.05 &+& 0.40z &=& 6.70 \\ & && && 0.40z &=& 6.70 -6.05 \\ & && && 0.40z &=& 0.65 \\ & && && z &=& \frac{0.65} {0.40}\\ & && && \mathbf{z} &\mathbf{=}& \mathbf{ 1.625 }\\ \\ \hline \\ (1) & && y && &=& 11 - 4z\\ & && y && &=& 11 - 4\cdot 1.625\\ & && y && &=& 11 - 6.5\\ & && \mathbf{y} && &\mathbf{=}& \mathbf{ 4.5 }\\ \\ \hline \\ x= 3z \\ x = 3\cdot 1.625\\ \mathbf{x = 4.875}\\ \end{array}\)

343000000000000000000000000000000 in standard form please. I got 3.43*10^32

\(3\ 43\ 000\ 000\ 000\ 000\ 000\ 000\ 000\ 000\ 000\ 000 = 3. 43 \cdot 10^{32}\)

Find all zeros of f(x)=2x^3+3x^2-23x-12

Find all zeros of f(x)=2x^3+3x^2-23x-12

\(2x^3+3x^2-23x-12=0\)

1. Try all divider of 12: \(\pm 1, ~ \pm 2,~ \pm 3,~ \pm4, ~\pm 6,~\pm12\)

We find \(x_1 = 3\) and \(x_2 = -4\)

2. 

\((x-3)(x+4) = x^2+4x-3x-12 = x^2+x-12\)

\(2x^3+3x^2-23x-12 : x^2+x-12=2x+1\)

\(2x_3+1 = 0\\ 2x_3 = -1\\ x_3 = -\frac{1}{2}\)

We have zeros  \(\mathbf{x_1=3 \qquad x_2 = -4 \qquad x_3= -\frac{1}{2}}\)

-5x - 10 = 10  x = ?

\(\begin{array}{rcll} -5x-10 &=& 10 \qquad | \qquad +10\\ -5x &=& 10 +10\\ -5x &=& 20 \qquad | \qquad \cdot(-1)\\ 5x &=& -20 \\ 5x &=& -20 \qquad | \qquad :5\\ x &=& -\frac{20}{5} \\ x &=& -\frac{4\cdot 5}{5}\\ \mathbf{x} &\mathbf{=}& \mathbf{-4}\\ \end{array}\)

17+42/53-98

17+42/53-98 = -80.2075471698113208

The latitude value of Toronto is 43.70 degrees, and the longitude value is 79.40 degrees. The latitude value of Melbourne is -37.81 degrees, and the longitude value is 144.96 degrees. The two cities are degrees apart in latitude. The two cities are degrees apart in longitude.

Sorry my delta longitude is false.

\(\begin{array}{lcl} \Delta \lambda &=& 144.96^{\circ}E + 79.40^{\circ}W = 224.36^{\circ} \\ \Delta \varphi &=& 43.70^{\circ} -(-37.81^{\circ} ) = 43.70^{\circ} +37.81^{\circ} =81.51^{\circ} \end{array}\)

The two cities are degrees apart in latitude \(81.51^{\circ}\)

The two cities are degrees apart in longitude \(224.36^{\circ}\)

1/12+7/9=63/72 ?

\(\begin{array}{rcl} \frac{1}{12} + \frac{7}{9}\\ &=& \frac{1}{12}\cdot\frac{3}{3} + \frac{7}{9}\cdot\frac{4}{4}\\ &=& \frac{3}{36} + \frac{28}{36}\\ &=& \frac{31}{36}\cdot\frac{2}{2}\\ &=& \frac{62}{72}\\ \end{array}\)